In the book called "Evaluating Feynman Integrals" by Smirnov, the following formula is given for partial fraction decomposition (p.35, eq. (3.10)):
$$\frac{1}{(x+x_1)^{a_1}(x+x_2)^{a_2}}=\sum_{i=0}^{a_1-1} \binom{a_2 - 1+i}{a_2} \frac{(-1)^i}{(x_2-x_1)^{a_2+i}(x+x_1)^{a_1-i}}+\sum_{i=0}^{a_2-1} \binom{a_1 - 1+i}{a_1} \frac{(-1)^i}{(x_2-x_1)^{a_1+i}(x+x_1)^{a_2-i}} \tag{1}$$
The author makes the following comment below (p.35):
If one of the indices, e.g. $a_2$ is non-positive, a similar decomposition is performed by expanding $(x+x_2)^{-a_2}$ in powers of $x +x_1$.
I am facing such a case, but I cannot really understand what is meant by the expansion. I could do a Taylor expansion I guess:
$$\begin{align}\frac{1}{(x+x_1)^{a_1}(x+x_2)^{a_2}} &=\frac{1}{(x+x_1)^{a_1}} \sum_{k=0}^{\infty} \frac{(x+x_1)^k }{k!}\frac{\partial^k}{\partial x^k} (x+x_2)^{-a_2} \rvert_{x=-x_1} \\ &= \sum_{k=0}^{\infty} \frac{(x+x_1)^{k-a_1} }{k!}\frac{\partial^k}{\partial x^k}(x+x_2)^{-a_2} \rvert_{x=-x_1} \tag{2}\end{align}$$
But now the power of $x+x_1$ is negative for $k<a_1$ and positive otherwise, which complicates the situation in the case where an integral must be performed afterwards. The infinite sum is also not so great. Moreover, the author says to do a "similar decomposition", which seems unneeded here. How can I do a decomposition in the case of $a_2<0$?
if $a_2 < 0$, letting $n = -a_2 > 0$, what you are expanding is $$\frac {(x + x_2)^n}{(x+x_1)^{a_1}}$$
The idea the author is expressing is that if we let $u = x + x_1$, then $x + x_2 = u + b$ where $b = x_2 - x_1$. And we can rewrite the expression as $$\frac {(u + b)^n}{u^{a_1}} = \sum_{i=0}^n{n\choose i}u^{i-a_1}b^{n-i} \\=(x_2-x_1)^{-a_2}\sum_{i=0}^{-a_2}{n\choose i}\frac 1{(x - x_1)^{a_1 - i}(x_2-x_1)^i}$$
To answer your question in the comment below, where $x, x_1, x_2$ are vectors in a real inner product space, where $(x+x_i)^{a_i}$ is replaced by $\|x + x_i\|^{a_i}$.
Again, let $u = x+x_1, b = x_2 - x_1$, and now let $n = -a_2/2, m = a_1/2$. We can now rewrite the expression as $$\frac{\langle u + b, u+b\rangle^n}{\langle u,u\rangle^m}= \frac {\left(\langle u,u\rangle + 2\langle u,b\rangle + \langle b,b\rangle\right)^n}{\langle u,u\rangle^m}$$ This expands to $$\sum_{k=0}^n\sum_{i=0}^k\binom n k\binom k i 2^{k-i}\langle u,u\rangle^{i-m}\langle u,b\rangle^{k-i}\langle b,b\rangle^{n-k} = \sum_{i+j+k = n} \frac{n!}{i!j!k!}2^j\langle u,u\rangle^{i-m}\langle u,b\rangle^j\langle b,b\rangle^k$$
However here I do not see a way to get all dependency on $u$ (and therefore on the variable $x$) in the denominator. I'm not sure what use this would be for your purposes.